Question

A gas is compressed in a piston-cylinder assembly from P1= 2 bar to P2= 8 bar, V2= 0.02 m^3 in a process during which the relation between pressure and volume is PV^1.3= Constant. the mass of the gas is 0.2 kg. If the specific internal energy of the gas increases by 50 kJ/kg during the process, determine the heat transfer, in kJ. Kinetic and Potential energies are negligible.

Answer 1

heat transfer is 4.6067 kJ

Step-by-step explanation:

given data

P1= 2 bar = 2 ×
`10^(5)` N/m²

P2= 8 bar = 8 ×
`10^(5)` N/m²

V2= 0.02 m³

m = 0.2 kg

`PV^(2=1.3)` = constant

V(1 to 2) = 50 kJ/kg

U(1 to 2) = m V(1 to 2) = 0.2 × 50 = 10 kJ

solution

`P_1 V_1^(1.3) = P_2 V_2^(1.3)`

`2*10^5 V_1^(1.3) = 8*10^5 * 0.02^(1.3)`

v1 = 0.05809 m³

and

by 1st law of thermodynamics for closed system is

Q - W = ΔU

so calculate the work expansion that is

W =
`(P_1V_1 - P_2 V_2)/(n-1)`

W =
`(2*10^5 * 0.05809 - 8*10^5 * 0.02)/(1.3-1)`

W = -14.6067 kJ

so heat transfer from 1st law of thermodynamic is

Q = U + W

Q = 10 + ( - 14.06067 )

Q = -4.6067kJ

so

heat transfer is 4.6067 kJ